Proof: Existence of Linearly Independent Vectors in Images of Non-scalar-multiple Linear Transformations

• by Yuehao Wang
• Mar 17, 2020

This is a challenging problem in an exercise of SI131 Linear Algebra for Information Science. This article elaborates my thoughts in this problem.

Proposition: Let $\tau_1, \tau_2: V\rightarrow V$ be linear transformations such that one is not a scalar multiple of the other. Suppose that $dim~Im(\tau_1),dim~Im(\tau_2)\geq 2$.

Then there exists a $v\in V$ such that $\tau_1(v), \tau_2(v)$ are linearly independent.

Proof:

Let $Im(\tau_1)$ be a subspace spanned by a set of basis $U$, $Im(\tau_2)$ be a subspace spanned by a set of basis $W$. There are two cases of $U$ and $W$:

1. If $\exists u_a \in U$, $u_a$ is linearly independent to basis in $W$.

   Find a $v_1, v_2\in V$ such that $\tau_1(v_1)=u_a$, $\tau_2(v_2)$ and $\tau_2(v_1)$ are not both zero (thus $\tau_2(v_2) + \tau_2(v_1)$ is not zero). We can guarantee that $u_a + \tau_1(v_2) \neq 0$ because we can arbitrarily scale $v_2$ and $\tau_2(v_2)+\tau_2(v_1) \neq 0$ will not be violated. Also notice that we can always find such $v_1$ and $v_2$ since $dim~Im(\tau_1), dim~Im(\tau_2) \geq 2$.

   Then we construct a new vector $v_1+v_2 \in V$,

   $\tau_1(v_1+v_2)=u_a + \tau_1(v_2) \neq 0$

   $\tau_2(v_1+v_2)=\tau_2(v_1)+\tau_2(v_2) \neq 0$

   1) If $\tau_2(v_1) \neq 0$ and $\tau_2(v_2) \neq 0$. Due to $u_a$ is linearly independent to any basis of $Im(\tau_2)$, $u_a$ is linearly independent to $\tau_2(v_1)$ and $\tau_2(v_2)$ $\Rightarrow$ $\tau_1(v_1+v_2)$ and $\tau_2(v_1+v_2)$ are linearly independent.

   2) If $\tau_2(v_1) = 0$ $\Rightarrow$ $\tau_2(v_2) \neq 0$. $u_a$ is linearly independent to $\tau_2(v_2)$ $\Rightarrow$ $\tau_1(v_1+v_2)$ and $\tau_2(v_1+v_2)$ are linearly independent.

   3) If $\tau_2(v_2) = 0$ $\Rightarrow$ $\tau_2(v_1) \neq 0$. Similar to 2).

2. If $\forall u_a \in U$, $u_a$ is linearly dependent to basis in $W$, then $Im(\tau_1)$ is also spanned by a set of basis $W$, i.e. $Im(\tau_1), Im(\tau_2)$ share a subspace.

   Hypothesis: Suppose $\forall v \in V$, $\tau_1(v), \tau_2(v)$ are linearly dependent. Therefore, $\forall v_1, v_2 \in V$, we have:

   $\tau_1(v_1) = k_1\tau_2(v_1)$

   $\tau_1(v_2) = k_2\tau_2(v_2)$

   W.L.O.G, we only care those cases in which $\tau_1(v_1)$ and $\tau_1(v_2)$ are non-zero. We can always find these $v_1$ and $v_2$ because $dim~Im(\tau_1), dim~Im(\tau_2)\geq 2$.

   Then we construct a vector $v_1 + v_2 \in V$.

   1) If $\tau_1(v_1)$ and $\tau_1(v_2)$ are linearly dependent and $\tau_1(v_1), \tau_1(v_2) \neq 0$. Let $\tau_1(v_2)=c\tau_1(v_1)$, we have:

   $\tau_1(v_1+v_2) = (1+c)\tau_1(v_1)$

   $\tau_2(v_1+v_2)=\frac{1}{k_1}\tau_1(v_1)+\frac{1}{k_2}\tau_1(v_2)=(\frac{1}{k_1}+\frac{c}{k_2})\tau_1(v_1)$

   Since they are still linearly dependent according to the hypothesis, $\exists k_3$:

   $[k_3(1+c) + (\frac{1}{k_1}+\frac{c}{k_2})]\tau_1(v_1)=0$

   Thus, we have $k_1 = k_2 = -1/k_3$.

   2) If $\tau_1(v_1)$ and $\tau_1(v_2)$ are linearly independent.

   $\tau_1(v_1+v_2) = k_1\tau_2(v_1) + k_2\tau_2(v_2)$

   $\tau_2(v_1+v_2)=\tau_2(v_1) + \tau_2(v_2)$

   Since they are still linearly dependent according to the hypothesis, $\exists k_3$:

   $(k_3+k_1)\tau_2(v_1)+(k_3+k_2)\tau_2(v_2)=0$

   Due to the independence of $\tau_1(v_1)$ and $\tau_1(v_2)$, $k_3+k_1 = k_3+k_2 = 0$.

   Thus, we have $k_1=k_2=-k_3$.

   3) For $\forall v\in Ker(\tau_1), u\in V$, we have:

   $\tau_1(v+u)=\tau_1(u)$

   $\tau_2(v+u)=\tau_2(v) + \tau_2(u)$

   If $\tau_1(u)$ is linearly independent to $\tau_2(v)$, $\tau_1(v+u)$ and $\tau_2(v+u)$ can not be linearly dependent (contradict to our hypothesis). Thus, $\tau_1(u)$ is always linearly dependent to $\tau_2(v)$, even if we choose any basis as $\tau_1(u)$. In order to meet our condition $dim~Im(\tau_1)\geq 2$, $\tau_2(v)=0$, i.e. $Ker(\tau_2)\subset Ker(\tau_1)$. By the symmetric argument on $\tau_2$, we have $Ker(\tau_1)\subset Ker(\tau_2)$.

   Thus, we have $Ker(\tau_1) = Ker(\tau_2)$.

   1), 2), 3) $\Rightarrow$ for any $v \in V$, if $\tau_1(v), \tau_2(v)$ are linearly dependent, then $\tau_1(v)=k_1\tau_2(v)$.

   However, $\tau_1, \tau_2$ are linear transformations such that one is not a scalar multiple of the other. $\Rightarrow$ Contradiction, i.e. $\exists v\in V$ such that $\tau_1(v)$ and $\tau_1(v)$ are linearly independent.